anonymous
  • anonymous
The x co-ordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
9y^2=x^3 d(9y^2)/dx =d(x^3)/dx 18y dy/dx =3x^2 dy/dx =x^2/(6y)
anonymous
  • anonymous
The gradient of tangent is x^2/(6y) So, gradient of normal is -6y/(x^2)
anonymous
  • anonymous
NOW, let x-intercept of the normal be a then y-intercept is also a.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Gradient of normal = (a-0)/(0-a) -6y/(x^2) =-1
anonymous
  • anonymous
Now,|dw:1359201825424:dw| and solve for x
anonymous
  • anonymous
|dw:1359201971611:dw|
anonymous
  • anonymous
U need to solve that
anonymous
  • anonymous
|dw:1359202095063:dw|
anonymous
  • anonymous
9y^2=x^3
anonymous
  • anonymous
-6y/(x^2) =-1......then wat did u do ?
anonymous
  • anonymous
Substitute y interms of x

Looking for something else?

Not the answer you are looking for? Search for more explanations.